# What variables does the action $S$ depend on?

Action is defined as,

$$S ~=~ \int L(q, q', t) dt,$$

but my question is what variables does $S$ depend on?

Is $S = S(q, t)$ or $S = S(q, q', t)$ where $q' := \frac{dq}{dt}$?

In Wikipedia I've read that $S = S(q(t))$ and I think that suppose, $q$ and $t$ are considered as independent coordinates. Then $S$ should depend on $q'$ also because, for the typical lagrangian

$$L ~=~ \frac{q'^2}{2} - V(q).$$

I think we cannot find a function $Z(q, t)$ such that $$L = \frac{dZ(q, t)}{dt}?$$

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If you like this question you may also enjoy this Phys.SE post. –  Qmechanic Oct 2 '11 at 21:16

1) Firstly, the Lagrangian $L(q(t),v(t),t)$ at some time $t$ is a function of:

1. the instantaneous position $q(t)$ at the time $t$;
2. the instantaneous velocity $v(t)$ at the time $t$; and
3. the time $t$ (also known as explicit time-dependence).

2) Secondly, the (off-shell) action

$$\tag{1} S[q]~:=~ \left. \int_{t_i}^{t_f}\! dt \ L(q(t),v(t),t)\right|_{v(t)=\dot{q}(t)}$$

is a functional of the full position curve/path $q:[t_i,t_f] \to \mathbb{R}$ for all times $t$ in the interval $[t_i,t_f]$.

3) Thirdly, if one imposes boundary conditions (B.C.), e.g. Dirichlet B.C.,

$$\tag{2} q(t_i)~=~q_i \qquad \text{and}\qquad q(t_f)~=~q_f,$$

then there is also a notion of a (Dirichlet) on-shell action $^1$

$$\tag{3} S(q_f,t_f;q_i,t_i)~:=~S[q_{\rm cl}]$$

where $q_{\rm cl}:[t_i,t_f] \to \mathbb{R}$ is the classical path, which satisfies Euler-Lagrange equations with the Dirichlet B.C. (2). The on-shell action $S(q_f,t_f;q_i,t_i)$ is a function of

1. the initial time $t_i$;
2. the initial position $q_i$;
3. the final time $t_f$; and
4. the final position $q_f$.

4) To answer the last subquestion(v2): The Lagrangian is in general not a total time-derivative, cf. this Phys.SE post.

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$^1$ See also e.g. MTW Section 21.1. For the on-shell action $S(q_f,t_f;q_i,t_i)$ to be well-defined, there should exist a unique classical path with the B.C. (2). (Here the words on-shell and off-shell refer to whether the Euler-Lagrange equations are satisfied or not.)

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Why do you add at v(t) = qdot(t)? instead, why not just write qdot(t) in place of v(t) and write L(q, qdot, t)? –  user5198 Oct 3 '11 at 14:53
@user5198: You can do that; it is equivalent. –  Qmechanic Oct 3 '11 at 16:54